Expert Answers to Algebra 2 Problems

Infinite inequality with logarithm
Solve inequality $3-<!--\; -\; -->{\log }_{0.5}<!--\; \; -->x-<!--\; -\; -->{\log }_{0.5}^2<!--\; \; -->x-<!--\; -\; -->{\log }_{0.5}^3<!--\; \; -->x-<!--\; -\; -->\cdots <!--\; \cdots \; -->\ge <!--\; \ge \; -->4{\log }_{0.5}<!--\; \; -->x$ Any suggestions how to start?

Basic calculus question about epidemic modeling
I am a medical student attempting to build mathematical models of disease, but am struggling with the calculus. In particular, I am very confused by this equation shown in attached photo, and feel worse because it is called "elementary". How is the second equation the integral of the first?
Feel free to ignore the first sentence, before "we consider."
The assumption (i) requires a fuller mathematical explanation, since the assumption of a recovery rate proportional to the number of infectives has no clearepidemiological meaning. We consider the "cohort" of members who were all infected at one time and let u(s) denote the number of these who are still infectivea time units after having been infected. If a fraction a of these leave the infective class in unit time then
$u^{\prime }=\mathrm{\alpha <!--\; \alpha \; -->}u,$
and the solution of this elementary differential equation is
$u(s)=u(0)e^{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}s}$

Potential such that the time-independent Schroedinger equation has an explicit solution
Consider the time-independent Schroedinger equation
${\mathrm{\varphi <!--\; \varphi \; -->}}^{″}(x)+V(x)\mathrm{\varphi <!--\; \varphi \; -->}(x)=\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\varphi <!--\; \varphi \; -->}(x)$
For testing my numerics, I would like to now:
For which choices of V are there explicit expressions for the solutions $(\mathrm{\varphi <!--\; \varphi \; -->},\mathrm{\lambda <!--\; \lambda \; -->})$?
For example, if $V(x)=x^2$, then the solutions are products of Hermite polynomials and $exp(-<!--\; -\; -->x^2)$ (up to correct scales) and arithmetic progressions (the same works in higher dimensions with the analogue potential).

Modeling drunkeness over time
So you have the function of drunkenness,D(t), the amount of alcohol in your blood, for $t>0$, over time,t in hours:
$D(t)=T-<!--\; -\; -->ct$
c being the arbitrary amount of alcohol your liver will process in 1 hour. T being the total amount of alcohol you have drunk at time t.
So T does depend on t, but it still a totally arbitrary amount.
As a piece wise function based on $T=c_{n}t$. $c_n=nc$ with n being some non negative number, representing the number of drinks of strength c had in 1 hours time:
$T=\begin{array}{cc}\{& \begin{array}{cc}t=1& n_{1}c\\ t=2& n_{2}c\\ ...\\ t=n& n_{n}c\end{array}\end{array}$
So to find T for a certain time, t,
$T(t=n)=n_{1}c+n_{2}c...+...n_{n}c=(n_1+n_2+n_n)c$
The number of drinks had each hour is totally arbitrary. Now what I want to do is to compare this between different values of c. Like comparing the same amount of drinking between a light beer (call it 1c) and a craft beer (2c). Is there a way to generalize this.
For example:
$T=\begin{array}{cc}\{& \begin{array}{cc}t=1& 2c\\ t=2& 1c\\ t=3& 3c\end{array}\end{array}$
$T(t=3)=6c$
So
$D(t=3)=3c$
but if $c=2c$
$T(t=3)=12c$
and
$D(t=3)=9c$
So drunkenness is not a linear relationship related to the strength of alcohol you are drinking.

Why does the branch cut for log(1+z), cutting away the negative axis, start at -1?
I am really confused about this concept.
Is it because from -1 to 0, the inputs (1+z) are just viewed as inputs that aren't actually from the negative axis at all, and so log(1+z) is well-defined and continuous on (-1,0]? I don't see the single-valuedness on (-1,0], though. Shouldn't log(1+z) jump by a $2\mathrm{\pi <!--\; \pi \; -->}i$ term, on (-1,0]?
Any comments are welcome.
Thanks,

How to prove "Logarithms grow more slowly than polynomials"
"Logarithms grow more slowly than polynomials. That is, $\Theta (\lg <!--\; \; -->n)$ grows more slowly than $\frac{-<!--\; -\; -->b\pm <!--\; \pm \; -->\sqrt{b^2-<!--\; -\; -->4ac}}{2a}$ for any positive constant a."
if $y_1=x^{1/2}$) and $y_2=\log <!--\; \; -->(x)$
by comparing the graphs can say $y_1>y_2$
But, is there a way to prove this?

logarithmic Series
I'm aware that by properties of logarithm
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}\ln <!--\; \; -->(k)=\ln <!--\; \; -->(n!)$
My question is if
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)={\ln }^2<!--\; \; -->(n!)?$
Because when I am verifying the value where $n=5$, I get different result... maybe Im missing something... is there a formula that defines
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)\text{ ?}$
Here's my computation:
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)={\ln }^2<!--\; \; -->(n!)$
$\underset{k=1}{\overset{5}{\sum <!--\; \sum \; -->}}{\ln }^2<!--\; \; -->(k)={\ln }^2<!--\; \; -->(5!)$
${\ln }^2<!--\; \; -->(1)+{\ln }^2<!--\; \; -->(2)+{\ln }^2<!--\; \; -->(3)+{\ln }^2<!--\; \; -->(4)+{\ln }^2<!--\; \; -->(5)={\ln }^2<!--\; \; -->(120)$
$0+0.48045+1.20694+1.92181+2.59029=22.92007$
$6.199494$ is not equal to $22.92007$

I need help solving a logarithm equation.
these were my steps. Can someone tell me where have i goe wrong since the answer is $-<!--\; -\; -->\frac{1}{4}$
$\sqrt{\log <!--\; \; -->(\sqrt{10}a)}=\frac{1}{2}$
$\frac{1}{2}\log <!--\; \; -->(\sqrt{10}a)=\log <!--\; \; -->\sqrt{10}$
${\log <!--\; \; -->(\sqrt{10}a)}^{\frac{1}{2}}=\log <!--\; \; -->\sqrt{1}0$ because $({\log }_a<!--\; \; -->x^n=n{\log }_a<!--\; \; -->x)$
$(\sqrt{10}a{)}^{\frac{1}{2}}={10}^{\frac{1}{2}}$ (squaring)
${10}^{\frac{1}{2}}a=10$ division by ${10}^{\frac{1}{2}}$
$a={10}^{1-<!--\; -\; -->\frac{1}{2}}$
$a=\sqrt{10}$

Limit evaluate $\underset{x\to <!--\; \to \; -->0}{lim}\frac{\ln <!--\; \; -->(\cos <!--\; \; -->(-<!--\; -\; -->5x))}{\ln <!--\; \; -->(\cos <!--\; \; -->(-<!--\; -\; -->3x))}$
Please help me with this limit without using L'Hôpital's rule. I would by happy if you use simple solving. Thank you as much as I can ;).

Determine which of the following differential equations are separable. Do not solve the equations. y'-xy=0

How to compute $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\displaystyle \frac{\ln <!--\; \; -->(1+e^{\mathrm{\alpha <!--\; \alpha \; -->}x})}{\ln <!--\; \; -->(1+e^{\mathrm{\beta <!--\; \beta \; -->}x})}}$?
Where $\alpha >0$ and $\beta >0$

Calculus - limit of a function with logarithms
Compute
$\underset{x\to <!--\; \to \; -->0_{+}}{lim}\ln <!--\; \; -->(x\ln <!--\; \; -->a)\ln <!--\; \; -->\left(\frac{\ln <!--\; \; -->(ax)}{\ln <!--\; \; -->(\frac{x}{a})}\right)$
where $a>1$
I am trying to get to a result withou using any advanced methods or even things such as l'Hospital's rule etc..
I got to a phase where I took the limit of the first logarithm which we can see tends to 0 from rewriting it as $\ln <!--\; \; -->a^x$. Then I wanted to make some adjustments to the second part of the expression and I got to the stage where I have the limit of
$(\ln <!--\; \; -->a^2)\left(\frac{x-<!--\; -\; -->a}{a}\right)$
That wont give me exact result but I should be able to justify that it the expression is defined and by multiplying it with the first limit which is $0$, the result should also be $0$
Can somebody please tell me how correct or wrong I am? Thanks.

If a polynomial is evaluated over an interval (by the use of interval arithmetic and Horner's method), what does the result say about the polynomial's roots?
I mean, if the resulting interval contains zero, does it mean there must be one more more roots inside the original interval? Could it happen that there might be no zero?
If the resulting interval contains no zero (i.e. both of its end-points have the same sign), does it mean there is no polynomial root inside the original interval? What if one of the end-points of the result is zero?

A polynomial whose roots form an arithmetic progression
Let f be a fourth degree polynomial whose roots form an arithmetic progression. Prove that f′'s roots also form an arithmetic progression.
I didn' t make much progress, I just wrote $f(x)=a(x-<!--\; -\; -->b-<!--\; -\; -->r)(x-<!--\; -\; -->b-<!--\; -\; -->2r)(x-<!--\; -\; -->b-<!--\; -\; -->3r)(x-<!--\; -\; -->b-<!--\; -\; -->4r)$ and I tried to differentiate, which obviously doesn't help too much.

What is the remainder on division of $z^{400}+z^{303}+1$ by $z^4-<!--\; -\; -->1$?
I am asked to determine the remainder when the polynomial $f(z)=z^{400}+z^{303}+1$ is divided by the polynomial $g(z)=z^4-<!--\; -\; -->1$.
I expressed f as $f(z)=h(z)(z^4-<!--\; -\; -->1)+r(z)$ where r(z) is a polynomial
Realising that $z^4-<!--\; -\; -->1=(z^2-<!--\; -\; -->1)(z^2+1)=(z-<!--\; -\; -->1)(z+1)(z^2+1)$, I expressed f as
$f(z)=h(z)(z-<!--\; -\; -->1)(z+1)(z^2+1)+r(z)$
I also wrote f as $f(z)=q(z)(z-<!--\; -\; -->1)+3$, since $f(1)=3$, and q is some polynomial.
So I had $f(z)=q(z)(z-<!--\; -\; -->1)+3=h(z)(z-<!--\; -\; -->1)(z+1)(z^2+1)+r(z)$
Hence, for some constant k, $q(z)=h(z)(z+1)(z^2+1)+k$
We know $f(-<!--\; -\; -->1)=1$
From $f(z)=q(z)(z-<!--\; -\; -->1)+3$, we have $f(-<!--\; -\; -->1)=q(-<!--\; -\; -->1)(-<!--\; -\; -->2)+3=1$, which implies $q(-<!--\; -\; -->1)=1$.
Then $q(-<!--\; -\; -->1)=h(-<!--\; -\; -->1)(0)+k=1$. Thus, $k=1$
We then have $q(z)=h(z)(z+1)(z^2+1)+1$, and from this, we have $f(z)=q(z)(z-<!--\; -\; -->1)+3=(h(z)(z+1)(z^2+1)+1)(z-<!--\; -\; -->1)+3$. Which simplifies to $f(z)=h(z)(z^4-<!--\; -\; -->1)+z+2$
And gives $r(z)=z+2$.
But I am given the answer is $r(z)=z^3+2$. What is going wrong here? Can I use modular arithmetic instead? How?

Evaluation of polynomial modulo in GF(2)
I'm new to this forum and I have a question about about modular arithmetics. My background is in computational chemistry and I work now as a software developer. For my work I need to learn about cryptography and I'm reading the book "Cryptography Made Simple" by Nigel P. Smart.
In paragraph 1.2 on Finite Fields, the author gives an example of polynomial modulo in ${\mathbb{F}}_2(x)/f(x){\mathbb{F}}_2(x)$ as such:
$(1+x+x^2)(x+x^3)(\mathrm{mod}{x}^4+1)=x^2+1$
However, when I do the evaluation myself I get a different result, which is confirmed by using the PolynomialMod function in Mathematica. I will show the steps I took below. First of all I expand the multiplication and evaluate modulo 2:
$x^5+x^4+2x^3+x^2+x(\mathrm{mod}2)=x^5+x^4+x^2+x$
Then:
$x^5+x^4+x^2+x(\mathrm{mod}{x}^4+1)$
$x(x^4+1)+x^4+x^2(\mathrm{mod}{x}^4+1)$
$x(x^4+1)+(x^4+1)+x^2-<!--\; -\; -->1(\mathrm{mod}{x}^4+1)$
Which finally results in:
$(x+1)(x^4+1)+x^2-<!--\; -\; -->1(\mathrm{mod}{x}^4+1)=x^2-<!--\; -\; -->1$
I put in attachment a screenshot 1 from Mathematica to confirm it matches my result. Initially I thought it was an error in the book, but in a following section the author gives another example:
$(x^3+1)(x^4+1)(\mathrm{mod}{x}^7+x+1)=x^4+x^3+x$
Again, evaluating the expression manually I obtain:
$(x^3+1)(x^4+1)(\mathrm{mod}{x}^7+x+1)=x^4+x^3-<!--\; -\; -->x$
Which is confirmed once more by Mathematica, see screenshot 2.
Does anyone know if there is a reason behind the difference in the results? Any help would be much appreciated.

The grade appeal process at a university requires that a jury be structured by selecting six individuals randomly from a pool of nine students and eight faculty.​ (b) What is the probability of selecting a jury of all​ faculty?

When can I use the natural log to help solve an integral?
Why is it okay to do this: $\int <!--\; \int \; -->\frac{1}{x-<!--\; -\; -->2}dx=\ln <!--\; \; -->(x-<!--\; -\; -->2)$
but not this: $\int <!--\; \int \; -->\frac{1}{1-<!--\; -\; -->x^2}dx=\ln <!--\; \; -->(1-<!--\; -\; -->x^2)$

Simplify. Assume that variable a is nonzero.
$(2a^{-<!--\; -\; -->4}{)}^{-<!--\; -\; -->4}$

Can all programs be modeled as operations of elementary arithmetic operations on inputs?
In mathematics and computabiltiy theory, we treat all inputs and intermediate results and final outputs as natural number. While algorithms/programs themselves are considered natural numbers, here we treat these programs/functions/algorithms as just computable functions.
The question is, when the function operates on an input to produce an output, can we consider the operation of function as using only a number of arithmetic operations (addition, subtraction, multiplication and division) on an input? Or does the use of if/else make the aforementioned not true?
If this is true, is the number of arithmetic operations polynomially proportional to the lowest time complexity bound possible for solving a problem? (That is, if the lowest time complexity is O(whatever), then the number of arithmetic operations is ${\text{O(whatever}}^k)$ where k is some rational number.)